 # CBSE Class 10 Maths Chapter 6 - Triangles Objective Questions

In this Chapter 6- Triangles of CBSE Class 10 Maths, the topics discussed include Introduction to Triangles, Similarity of Triangles, Areas of Similar Triangles, Pythagoras Theorem and so on. An important part of unit 3- Geometry, chapter 6- Triangles have nine theorems in total. Meanwhile, having a clear understanding of the concepts, theorems and the problem-solving methods in this chapter are mandatory to score well in the Board examination of class 10 maths. Taking this into consideration along with the fact that the changed exam pattern will include more MCQs, we have compiled the CBSE Class 10 Maths Chapter 6 -Triangles Objective Questions for the students to solve. More practice will help them master the concepts.

### List of Sub-Topics Covered In Chapter

6.1 Areas of Similar Triangles (4 MCQs From The Topic)

6.2 Basic Proportionality Theorem (4 MCQs Listed From Topic)

6.3 Criteria for Similarity of Triangles (4 MCQs From The Topic)

6.4 Pythagoras Theorem (4 MCQs Listed From The Given Topic)

6.5 Similar Triangles (4 MCQs From The Topic)

The CBSE Class 10 Maths Objective Questions containing these topics will help the students to prepare well for the exams.

## Download CBSE Class 10 Maths Chapter 6 -Triangles Objective Questions Free PDF

### Areas of Similar Triangles

1. If △ ABC ~ △ DEF such that AB = 12 cm and DE = 14 cm. Find the ratio of areas of △ ABC and △ DEF.
1. 49/9
2. 36/49
3. 49/16
4. 25/49

Solution: We know that the ratio of areas of two similar triangles is equal

to the ratio of the squares.

Of any two corresponding sides,

area of △ ABC / area of △ DEF = (AB/DE) 2= (12/14) 2= 36/49

1. D and E are points on the sides AB and AC respectively of a △ABC such that DE || BC. Which of the following statement is true?
1. △ ADE ~ △ ABC
1. only (iii)
2. only (i)
3. only (i) and (ii)
4. all (i) , (ii) and (iii)

Solution: In △ ADE and △ ABC, we have

[Since DE || BC ∠ ADE = ∠ B (Corresponding angles)]

and, ∠ A = △ A [Common]

Therefore, (area of △ ADE / area of △ ABC) = (AD2/AB2)

1. In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm and area (ΔQOA) = 150 cm2, find the area of ΔPOB.

1. 233 cm2
2. 294 cm2
3. 300 cm2
4. 420 cm2

Solution: Consider Δ~QOA and Δ POB

QA || PB,

Therefore, ∠ AQO = ∠ PBO [Alternate angles]

∠ QAO = ∠ BPO  [Alternate angles]

and

∠ QOA = ∠ BOP [Vertically opposite angles]

Δs QOA ~ BOP  [by AAA similarity]

Therefore, (OQ/ OB) = (OA/OP)

Now, area (POB)/ area (QOA) = (OP) 2/ (OA) 2= 72/ 52

Since area (QOA) =150cm2

⇒area (POB) =294cm2

1. Two isosceles triangles have equal angles and their areas are in the ratio 16: 25. The ratio of corresponding heights is:
1. 4:5
2. 5:4
3. 3:2
4. 5:7

Solution: For similar isosceles triangles,

Area (Δ1) / Area (Δ2) = (h1)2 / (h2)2

(h1 / h2) = 4/5

### Basic Proportionality Theorem

1. In △ABC, AB = 3 and, AC = 4 cm and AD is the bisector of ∠A. Then, BD : DC is —
1. 9: 16
2. 4:3
3. 3:4
4. 16:9

Solution: The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides (It may be similar or may not depending on type of triangle it divides)

In △ABC

as per the statement AB/ AC= BD/DC i.e. a/b= c/d

So, BD/ DC= AB/AC= ¾

So, BD: DC = 3: 4

1. ABCD is a parallelogram with diagonal AC If a line XY is drawn such that XY ∥ AB.

BX/XC=? 1. (AY/AC)
2. DZ/AZ
3. AZ/ZD
4. AC/AY

Solution: In the Δ ABC,

AB ∥ XZ

AB ∥ XY

∴ BX/ XC= AY/YC…. (By BPT)….. (1)

In parallelogram ABCD,

AB ∥ CD

AB ∥ CD ∥ XZ

In the Δ ACD,

CD ∥ YZ

∴ AY/YC= AZ/ZD … (By BPT)…… (2)

From 1 & 2,

BX/XC= AY/YC= AZ/ZD

BX/XC= AZ/ZD

1. In ABC, Given that DE//BC, D is the midpoint of AB and E is a midpoint of AC.

The ratio AE: EC is ____.

1. 1: 3
2. 1:1
3. 2:1
4. 1:2

Solution: DE is parallel to BC

∠DAE = ∠ECF {Alternate angles}

∠BAC = ∠DAE

⇒ AD/DB= AE/EC (Basic Proportionality Theorem)

Since, D is midpoint of AB.

⇒ AE/EC= 1 /1

∴ AE: EC= 1:1

1. In ΔABC, AC = 15 cm and DE || BC. If AB/AD=3, Find EC.
1. 5cm
2. 10 cm
3. 2.5cm
4. 9cm

Solution: Given: DE∥BC

From basic proportionality theorem

⇒2AE=EC⇒AC=AE+EC−−−−−−− (1)

On substituting value of EC in (1), we get

15=3AE⇒5=AE⇒EC=10cm

### Criteria for Similarity of Triangles

1. △ ABC is an acute angled triangle. DE is drawn parallel to BC as shown. Which of the following are always true?

i) △ ABC ∼ △ ADE

iii) DE= BC/2 1. Only (i)
2. (i) and (ii) only
3. (i), (ii) and (iii)
4. (ii) and (iii) only

Answer: (B) (i) and (ii) Only

DE = BC/2 only if D and E are the mid points of AB and AC respectively. So this may not be true always.

1. The triangles ABC and ADE are similar Which of the following is true?

2. BC/BD=CE/DE
4. All of the Above

Solution: Since the given triangles are similar, the ratios of corresponding sides are equal.

1. If in △ CAB and △ FED, AB/ EF=BC/FD=AC/ED, then:
1. △ ABC∼△ DEF
2. △ CAB∼△ DEF
3. △ ABC∼△ EFD
4. △ CAB∼△ EFD

Solution: If two triangles are similar, corresponding sides are proportional.

Therefore, △ABC∼△EFD.

1. A tower of height 24m casts a shadow 50m and at the same time, a girl of height 1.8m casts a shadow. Find the length of the shadow of girl.
1. 3.75m
2. 3.5m
3. 3.25m
4. 3m

Solution: In △ABC and △DEC

∠ABC=∠DEC=90°

∠C=∠C (common)

Therefore, △ABC∼△DEC       [by AA similarity]

So, DE/AB=EC/BC

EC=DE × (BC/AB)

EC= 1.8× (50/24) ⇒EC=3.75 m ### Pythagoras Theorem

1. In the adjoining figure, if BC = a, AC = b, AB = c and   ∠CAB = 120°, then the correct relation is- 1. a2 = b2 +  c2 – bc
2. a2 = b2 +  c2 + bc
3. a2 = b2 +  c2 – 2bc
4. a2 = b2 +  c+ 2bc

Answer: (B)a2 = b2 +  c2 + bc

Solution: In △CDB,

BC2 =  CD2 + BD2      [By Pythagoras Theorem]

BC2 =  CD2 + (DA+AB)

BC2 =  CD2 + DA2 + AB2 + (2×DA×AB)         (i)

CD2 + DA2 = AC2      (ii)  [By Pythagoras Theorem]

Putting the values from (ii) and (iii) in (i), we get

BC2 =  AC2 + AB2 + (AC×AB)

a2 = b2 +  c2 + bc

Alternatively,

Since  ∠A is an obtuse angle in  ΔABC, so

BC2 =  AB2 +  AC2 + 2AB . AD

= AB2 +  AC2 + 2×AB× ½ ×AC

[∵ AD = AC cos  60∘ = 1/2AC]

= AB2 +  AC2 + AB × AC

a2 = b2 +  c2 + bc.

1. If the distance between the top of two trees 20 m and 28 m tall is 17 m, then the horizontal distance between the trees is :
1. 11m
2. 31m
3. 15m
4. 9m

Solution: Let AB and CD be two trees such that AB = 20 m, CD = 28 m & BD = 17 m Draw BE parallel to CD. Then, ED = 8 m.

By applying Pythagoras theorem:

BE2+DE2=BD2  =15m

∴ AC = BE = 15 m

1. In the figure △ABC is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

Solution: ∠A=∠A (common angle)

1. In a right △ABC, a perpendicular BD is drawn on to the largest side from the opposite vertex. Which of the following does not give the ratio of the areas of △ABD and △ACB? 1. (AB/AC)2
4. (BD/CB)2

Solution:

Consider  ΔABD and ΔACB: ∠BDA =  ∠ABC     [ 90°]

By AA similarity criterion, △ABD ~ △ACB

Hence,

ar (ΔABD)/ ar(ΔACB) = (AB/AC)2= (AD/AB)2= (BD/CB)2

### Similar Triangles

1. △ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm.△ DEF is similar to △ABC. If EF = 4 cm, then the perimeter of △DEF is –
1. 7.5 cm
2. 15cm
3. 30cm
4. 22.5cm

Solution: AB/DE= AC/DF=BC/EF=2/4=1/2

DE = 2×AB = 6 cm, DF = 2×AC = 5 cm

∴ Perimeter of △DEF = (DE + EF + DF) = 15 cm.

1. In △ ABC and △ DEF, ∠A = ∠E = 40∘ and AB/ED=AC/EF. Find ∠B if ∠F is 65°
1. 85°
2. 75°
3. 35°
4. 65°

Solution: AB/ED= AC/EF (Given)

∠A = ∠E = 40°

Since, the ratio of adjacent sides and the included angles are equal.

∴△ABC is similar to △EDF by SAS similarity criterion.

Now, ∠C = ∠F = 65°       [Corresponding angles of a similar triangles are equal]

∴ ∠B = 180°− (∠A+∠C)

=180° – (40° + 65°) = 75°

1. The ratio of the corresponding sides of two similar triangles is 1: 3. The ratio of their corresponding heights is _________
1. 1:3
2. 3:1
3. 1:9
4. 9:1

Solution: Ratio of heights = Ratio of sides = 1: 3.

1. If Δ ABC and Δ DEF are similar such that 2AB = DE and BC = 8 cm, then Find EF.
1. 16 cm
2. 12 cm
3. 8 cm
4. 4 cm

Solution: 2AB = DE

2BC = EF

⇒ 2×8 = EF

⇒ EF = 16 cm

## CBSE Class 10 Maths Chapter 6 Extra MCQ

1. A square and a rhombus are always_____
(a)congruent
(b) similar but congruent
(c) similar
(d) neither similar nor congruent

2. What is the length of each side of a rhombus whose diagonals have lengths 10 cm and 24  cm?
(a)13 cm
(b)26 cm
(c) 25 cm
(d)34 cm

Triangle is the most interesting and exciting chapters of unit 3- Geometry as it takes the students through the different aspects and concepts related to the Geometrical figure triangle. A triangle is a plane figure with three sides and three angles. This chapter deals with several topics and sub-topics related to triangles. Check out below for some sub-topics covered in this chapter.

Get more such CBSE class 10 Maths and Science notes at BYJU’S. Also, access class 10 question papers, sample papers, and other study materials to prepare for the board exam in a more effective way.